Confusion regarding the Q point of a transistor

Many standard text books define the Q point as follows :

"Q point or the operating point of a device, also known as a bias point, or quiescent point is the steady-state DC voltage or current at a specified terminal of an active device such as a diode or transistor with no input signal applied".

Now let me explain why it is confusing to me.

As we know the signal is supposed to apply at the base of the transistor, "with no input signal applied" means the base voltage is zero at some point, but that means the transistor is off at that point and as a consequence, there will not any collector current, missing something?

• The "signal" is the AC portion of the voltage. DC bias still present. Oct 22, 2020 at 15:53
• thanks, can you explain a bit more? Oct 22, 2020 at 16:00
• There is enough literature explaining it in detail. In a nutshell to bring a circuit to operation point you apply certain DC bias to its components to bring it to the linear region you are interested in. Then adding a small dc signal to this DC will make the circuit to work as if it was linear for this signal around this point. Oct 22, 2020 at 16:03
• "No signal" should be understood to mean "except the bias voltage, because that isn't a signal".
– user16324
Oct 22, 2020 at 18:19

As we know the signal is supposed to apply at the base of the transistor,

This is true for a common emitter or common collector stage, but for a common base stage, the signal will be applied at the emitter.

"with no input signal applied" means the base voltage is zero at some point,

This is not correct. "No signal" in this context means there are no variations about the Q point, or that the AC component of the input voltage is zero. It doesn't mean the instantaneous voltage is zero.

but that means the transistor is off at that point and as a consequence, there will not any collector current, missing something?

It doesn't. The collector current will be whatever is established by the DC bias at the Q point.

You can examine the distortion of a bipolar transistor under the combined DC_bias and the AC_peak_peak excursions.

If I recall rightly, 4 millivolts PeakPeak will cause 10% distortion.

Understanding this will be a big step forward in applying bipolars in your circuits.